1
2
3
47
6
Journal of Integer Sequences, Vol. 13 (2010),
Article 10.9.6
23 11
On the Extension of the Diophantine
√
Pair {1, 3} in Z[ d]
Zrinka Franušić
Department of Mathematics
University of Zagreb
Bijenička cesta 30
10000 Zagreb
Croatia
fran@math.hr
Abstract
√
In this paper, we consider Diophantine triples of the form {1, 3, c} in the ring Z[ d].
We prove that √the Diophantine pair {1, 3} cannot be extended to the Diophantine
quintuple in Z[ d] with d < 0 and d 6= −2.
1
Introduction
A Diophantine m-tuple in a commutative ring R with the unit 1 is a set of m distinct nonzero elements with the property that the product of each two distinct elements increased by
1 is a perfect square in R. These sets are studied in many different rings: the ring of integers
Z, the ring of rationals
√ Q, the ring of√Gaussian integers Z[i] ([2, 8]), the ring of integers
of quadratic fields Z[ d] and Z[(1 + d)/2] ([7, 9]), polynomial rings ([4, 5]). The most
1 33 17 105
, 16 , 4 , 16 } (found
famous Diophantine m-tuples and historical examples are quadruples { 16
by Diophant) and {1, 3, 8, 120} (found by Fermat). More information about these sets can
be found on Dujella’s web page on Diophantine m-tuples [3].
In 1969, Baker and Davenport [1] showed that the Diophantine triple {1, 3, 8} can be
extended uniquely to the Diophantine quadruple {1, 3, 8, 120} A030063 . Hence, the triple
{1, 3, 8} cannot be extended to a Diophantine quintuple. Jones [10] proved that if the set
{1, 3, c} is a Diophantine triple then c must be of the form
√
√
√
√
1
ck = ((2 + 3)(7 + 4 3)k + (2 − 3)(7 − 4 3)k − 4),
6
1
for some k ∈ N (A045899). It can be easily verified that sets {1, 3, ck , ck−1 } and {1, 3, ck , ck+1 }
are Diophantine quadruples. Indeed, ck ck+1 + 1 = t2k , where (tk ) is A051048. Moreover, the
Diophantine triple {1, 3, ck } can be extended to a quadruple only by the elements ck−1 and
ck+1 . This assertion is proved in 1998. by Dujella and Pethö [6]. A direct consequence of
their assertion is the fact that the pair {1, 3} cannot be extended to a Diophantine quintuple.
A natural question that may arise is can we extend
√ the pair√{1, 3} to a quintuple in a
+ b d : a, b ∈ Z} where d ∈ Z
larger ring. In this paper, we have chosen the ring Z[ d] = {a√
and
√ d is not a perfect square. So, we assume that {1, 3, a + b d} is a Diophantine triple in
Z[ d] and according to a definition it means that there exist integers ξ1 , η1 , ξ2 , η2 such that
√
√
(1)
a + b d + 1 = (ξ1 + η1 d)2
√
√ 2
3a + 3b d + 1 = (ξ2 + η2 d) .
(2)
Although, it seams that we have much more ’freedom’ in larger rings, the result for some
rings stays the same.
Theorem 1. Let d be a negative integer and d 6= −2.
√ The Diophantine pair {1, 3} cannot
be extended to a Diophantine quintuple in the ring Z[ d].
√
In certain rings Z[ d] for positive
√ d, Theorem 1 is not valid. For example {1, 3, 8, 120, 1680}
is the Diophantine quintuple in Z[ 8 · 1680 + 1].
√
In last√two sections, we describe the set of all Diophantine triples {1, 3, c} in Z[ −2]
and in Z[ d] for some positive integer d, where c is an integer and such that {1, 3, c} is not
a Diophantine triple in Z. The existence of such triples is related to solvability of certain
Pellian equations.
2
Main result
Proposition
2. Let d be a negative integer. If {1, 3, c} is a Diophantine triple in the ring
√
Z[ d], then c is an integer.
√
Proof. Let c = a + b d. There exist ξ1 , η1 , ξ2 , η2 ∈ Z such that
a + 1 = ξ12 + η12 d, b = 2ξ1 η1 , 3a + 1 = ξ22 + η22 d, 3b = 2ξ2 η2 .
(3)
Evidently, b must be an even number. By squaring and subtracting corresponding relations
in (3), we get
(a + 1)2 − db2 = x2 , (3a + 1)2 − d(3b)2 = y 2 ,
(4)
where x = ξ12 − η12 d and y = ξ22 − η22 d. By eliminating the quantities b and d from (4), we
obtain
(3x)2 − y 2 = (3a + 3)2 − (3a + 1)2 = 4(3a + 2),
(5)
It is clear, from (5), that 3x − y and 3x + y must be even. So, we have
3x − y = 2k c1 , 3x + y = 2l c2 ,
2
where k, l are positive integers and c1 , c2 are odd. Thus, we have
(3x − y)(3x + y) = 2k+l n,
where and c1 c2 = n. Furthermore, it follows that
y = 2l−1 c2 − 2k−1 c1 .
The expression (3a + 1)2 − y 2 can be given in terms of c1 and c2 :
(3a + 1)2 − y 2 =
=
=
=
(2k+l−2 n − 1)2 − (2l−1 c2 − 2k−1 c1 )2
4k+l−2 n2 − 2k+l−1 n + 1 − 4l−1 c22 + 2k+l−1 c1 c2 − 4k−1 c21
4k+l−2 n2 + 1 − 4l−1 c22 − 4k−1 c21
(4k−1 c21 − 1)(4l−1 c22 − 1).
On the other hand, (3a + 1)2 − y 2 = d(3b)2 (by (4)). Since |ci | ≥ 1 and k, l ≥ 1, we obtain
that d < 0 implies b = 0.
According to the previous
proposition, it may have sense to study Diophantine triples of
√
the form {1, 3, c} in Z[ d] such that c is an integer.
√
The √
triple {1, 3, c}, c ∈ Z, will be called a proper triple in Z[ d] if it is a Diophantine
triple
√in
√ Z[ d], but not a Diophantine triple
√ in Z. For instance, {1, 3, 161} is a proper triple
Z[ 2], {1, 3, −3} is a proper triple in Z[ −2], but {1, 3, 120} is not a proper triple in Z[ d]
(d 6= 1).
√
Let us, now, assume that {1, 3, c} is proper triple in Z[ d]. So, one of the following cases
may occur
c + 1 = ξ2
(6)
3c + 1 = dη 2 ,
c + 1 = dη 2
3c + 1 = ξ 2 ,
(7)
c + 1 = dη12
3c + 1 = dη22 .
(8)
Following proposition gives a connection between the extensibility of the Diophantine
pair {1, 3} and solvability of certain Pellian equations.
Proposition 3. The
√ Diophantine pair {1, 3} can be extended to a proper Diophantine triple
{1, 3, c} ⊂ Z in Z[ d] if and only if one of the following equations
3x2 − dy 2 = 2
(9)
x2 − 3dy 2 = −2
(10)
and
is solvable in Z.
3
Proof. ⇒ If {1, 3, c} is a proper triple and (6) is fulfilled, then by eliminating c, we get
3ξ 2 − dη 2 = 2. So, the equation (9) is solvable in Z.
Analogously, by eliminating c from (7), we obtain that the equation (10) is solvable in Z.
Finally, from (8) we have
d(η22 − 3η12 ) = −2,
and the only possibilities are:
(a) d = −1 and η22 − 3η12 = 2, which is impossible (because the x2 − 3y 2 = 2 is not solvable
in Z) ,
(b) d = 2 and η22 − 3η12 = −1, which is also impossible for the same reason,
(c) d = −2 and η22 − 3η12 = 1, which is possible because the Pell’s equation x2 − 3y 2 = 1 has
infinitely many solution. Note that the equation (9) is also solvable if d = −2, it has
a unique solution (0, 1).
⇐ Assume that (ξ, η) ∈ Z2 is a solution of the equation (9). Then the set {1, 3, ξ 2 − 1}
√
√
represents a proper triple in Z[ d]. Indeed, 3(ξ 2 − 1) + 1 = dη 2 = (n d)2 .
Similarly, if (ξ, η) ∈ Z2 is a solution of (10), then we get a proper triple {1, 3, dη 2 −1}.
In order to prove Theorem 1, we need a special case of Theorem 8 from [10].
Lemma 4 (Theorem 8, [10]). If {1, 3, c} is a Diophantine triple in Z, then c = ck for some
integer k ≥ 2 where
ck = 14ck−1 − ck−2 + 6, c1 = 8, c0 = 0.
By solving the recursion from Lemma 4 we get
√
√
√
√
1
ck = ((2 + 3)(7 + 4 3)k + (2 − 3)(7 − 4 3)k − 4), k ≥ 0
6
(11)
Proposition
5. Let d be a negative integer and d 6= −2. If {1, 3, c} is a Diophantine triple
√
in Z[ d], then c = ck for some positive integer k, where ck is given by (11).
Proof. According
to Proposition 2, c is an integer. Hence, {1, 3, c} is a triple in Z or a proper
√
triple in Z[ d]. Since the equations (9) and (10) are not solvable
√ in Z for d < 0 and d 6= −2,
Proposition 3 implies that {1, 3, c} is not a proper triple in Z[ d]. Hence, it is a Diophantine
triple in Z. Finally, from Lemma 4 we get that c = ck for some k.
Lemma 6 (Theorem 1, [6]). If {1, 3, ck , d} is a Diophantine quadruple in Z where ck is given
by (11) and k is a positive integer, then d ∈ {ck−1 , ck+1 }.
At this point we can conclude that Theorem 1 is proved. (It follows directly from Proposition 5 and Lemma 6.)
Since the extension of the pair {1, 3} is closely related to the solvability of the Pellian
equation x2 − 3y 2 = −2, our result can be understood as the following statement.
4
Corollary 7. Let d be a negative integer and d 6= −2. The equation
x2 − 3y 2 = −2
(12)
√
in Z[ d] has only real solutions.
√
√
Proof. Suppose that (ξ1 + η1 d, ξ2 + η2 d) is a solution of the equation (12). For
√
√
a + b d = (ξ2 + η2 d)2 − 1,
√
√
we obtain that {1, 3, a + b d} is a Diophantine triple u Z[ d]. Indeed,
√
√
√
3(a + b d) + 1 = 3(ξ2 + η2 d)2 − 2 = (ξ1 + η1 d)2 .
√
According to Proposition 5, there exists some k such that a + b d = ck . Immediately, we
conclude that η1 = η2 = 0.
3
The case d = −2
√
In this section, we will take a closer look at the Diophantine
√ triples {1, 3, c} in Z[ −2]. In
fact, we are able to describe all Diophantine triples in Z[ −2].
√
Proposition 8. If {1, 3, c} is a Diophantine triple in Z[ −2], then c ∈ {−1, ck , dk } for
some k ∈ N, where ck is given by (11) and
dk = −
√
√
1
(7 + 4 3)k + (7 − 4 3)k + 4
6
(13)
in
Proof. By Proposition 2, we have that c is an integer. Since the equation (9) is solvable
√
Z, Proposition 3 implies that {1, 3, c} can be a triple in Z or a proper triple in Z[ −2].
√ If
{1, 3, c} is a triple in Z, then c = ck from Lemma 4. If {1, 3, c} is a proper triple in Z[ −2],
then one of the cases (6) and (8) may occur. The case (6) implies that c = −1, because
(0, 1) is the unique solution of (9).
In the case of (8), c = (−2x2 − 1)/3 where x is a solution of the Pell’s equation
x2 − 3y 2 = 1.
(14)
All solutions of (14) are
xk+2 = 4xk+1 − xk , x1 = 2, x0 = 1,
for k ≥ 0. By solving the recursion we obtain that
√
√
1
xk = ((2 − 3)k + (2 + 3)k ),
2
and dk = (−2x2k − 1)/3.
Note that for k = 0, we have d0 = −1. Also, it is interesting that (−dk ) is A011922.
5
Remark 9. It is known that {1, 3, ck , ck+1 } is a Diophantine quadruple
√ in Z for k ≥ 1
(Lemma 6). The same can be proven for the set {1, 3, dk , dk+1 } in Z[ −2] for k ≥ 0. We
have
√
√
36(dk dk+1 + 1) = 66 + (7 − 4 3)2k+1 + (7 + 4 3)2k+1 +
√
√
√
√
16((7 − 4 3)k (2 − 3) + (2 + 3)(7 + 4 3)k )
√
√
= 64 + 16((2 + 3)2k+1 + (2 − 3)2k+1 ) +
√
√
((2 + 3)2k+1 + (2 − 3)2k+1 )2
= 4(x22k+1 + 8x2k+1 + 16) = 4(x2k+1 + 4)2 ,
where x2k+1 is a solution of x2 − 3y 2 = 1. It can be easily shown that x2k+1 ≡ 2 (mod 3)
(because xn+2 = 4xn+1 − xn , x0 = 1, x1 = 2), so
is an integer.
p
1
dk dk+1 + 1 = (x2k+1 + 4)
3
Further, the set {1, 3, ck , dl } is not a Diophantine quadruple for k ≥ 1 and l ≥ 0. Indeed,
it can be easily seen that ck dl + 1 is a negative√odd number (from
Lemma 4 it follows that
√
ck is even, and from (13) that dl < 0). Hence, ck dl + 1 6∈ Z[ −2].
4
√
Proper triples in Z[ d] for some d > 0
√
According to Proposition 3, the set {1, 3, c} ⊂ Z is a proper Diophantine triple in Z[ d] if
and only if one of the equations (9), (10) is solvable in Z.
√
Example 10. Determine all proper Diophantine triples {1, 3, c} in Z[ 10].
Note that d = 10 is the smallest positive integer, d 6= 1, such that the equation √
(9) is
solvable in Z. If (xn , yn ) is a solution of (9), then {1, 3, x2n − 1} is a proper triple in Z[ 10].
All solutions of (9) are
xn+2 = 22xn+1 − xn , x1 = 42, x0 = 2,
for n ≥ 0. Hence,
√
√
√
√
1
xn = ((6 + 30)(11 + 2 30)n + (6 − 30)(11 − 2 30)n ), n ≥ 0.
6
√
So, all proper triples {1, 3, c} in Z[ 10] are {1, 3, dn } where
√
√
1
(15)
dn = x2n − 1 = ((11 + 2 30)2n+1 + (11 − 2 30)2n+1 − 4), n ≥ 0.
6
√
√
√
Besides these triples, sets {1, 3,
√ 25 ± 8 10}, {1, 3, 160 ± 44 10}, {1, 3, 355 ± 112 10}
are also Diophantine
triples in Z[ 10]. The third element of these triples√is related√to the
√
solution in Z[ 10] of the equation √
x2 − 3y 2 =√−2. For instance, (6 + 2 10, 4 + 10) is
solution of x2 −√3y 2 = −2 and 25 + 8 10 = (4 + 10)2 − 1. More generally,
if X 2 − 3Y 2 = −2
√
and X, Y ∈ Z[ 10] then {1, 3, Y 2 − 1} is a Diophantine triple in Z[ 10].
6
√
Example 11. Determine all proper Diophantine triples {1, 3, c} in Z[ 2].
√
Since the equation (10) is solvable in Z, {1, 3, 2yn2 − 1} is a proper triple in Z[ 2] where
(xn , yn ) is a solution of (10). All solutions of (10) are
yn+2 = 10yn+1 − yn , y1 = 9, y0 = 1,
for n ≥ 0, i.e.,
√
√
√
√
1
yn = ((5 − 2 6)n (3 − 6) + (3 + 6)(5 + 2 6)n ).
6
√
So, all proper triples {1, 3, c} in Z[ 2] are {1, 3, dn } where
√
√
1
dn = 2yn2 − 1 = ((5 + 2 6)2n+1 + (5 − 2 6)2n+1 − 4).
(16)
6
√
√
2],
there
exist
triples
like
{1,
3,
11
±
8
2},
Also, because√
the equation (10) is solvable
in
Z[
√
{1, 3, 32 ± 20 2}, {1, 3, 161 ± 112 2}. Like in √Example 10, the third element of these
triples is obtained from the solution
(X, Y ) ∈ Z[ 2]2 of the equation x2 − 3y 2 = −2, i.e.,
√
2
{1, 3, Y − 1} is a triple in Z[ 2].
√
It is interesting to note that the expression (15) contains 11 + 2 30 (i.e. (11, 2)) which
√
is a fundamental solution of the Pell’s equation x2 − 30y 2 = 1. Similarly, in (16), 5 + 2 6
represents a fundamental solution of x2 − 6y 2 = 1.
In√what follows, we will give the complete set of proper Diophantine triples {1, 3, c}
in Z[ d] for positive d which satisfies certain conditions. This problem is related to the
solvability of the equation x2 − 3dy 2 = 6. For this equation, we have the following Nagell’s
result.
Lemma 12 (Theorem 11, [11]). Let D be a positive integer which is not a perfect square,
and let C 6= 1, −D be a square-free integer which divides 2D. Then if (x0 , y0 ) is the least
positive (fundamental) solution of the equation
x2 − Dy 2 = C,
then
√
1
x2 + Dy02 2x0 y0 √
(x0 + y0 D)2 = 0
+
D
|C|
|C|
|C|
(17)
is a fundamental solution of the related Pell’s equation x2 − Dy 2 = 1. All solutions of (17)
are
√
√ !n
xn + yn D
x0 + y0 D
p
p
=
,
|C|
|C|
where n is a positive odd integer.
7
Theorem 13. Let d be a positive integer such that neither d nor 3d are perfect squares and
such that one
√ of the equations (9), (10) is solvable in Z. If {1, 3, c} is a proper Diophantine
triple in Z[ d], d 6= 2, 10, then there exists nonnegative integer n such that
√
√
1
c = dn = ((ξ + η 3d)2n+1 + (ξ − η 3d)2n+1 − 4),
6
(18)
where (ξ, η) is a fundamental solution of the Pell’s equation
x2 − 3dy 2 = 1.
(19)
If d ∈ {2, 10}, then c = dn for some n ≥ 1.
Proof. First, assume that the equation (9) is solvable in Z. Hence, c = x′2 − 1, where (x′ , y ′ )
is a solution of (9). By Lemma 12, we get that all solutions of (9) are
√
√
√
√
1
xn = ((α + β 3d)(ξ + η 3d)n + (α − β 3d)(ξ − η 3d)n ), n ≥ 0,
6
where (α, β) is a fundamental solution of the equation
x2 − 3dy 2 = 6
(20)
and (ξ, η) is a fundamental solution of the related Pell’s equation (19). (Note that (20) has
exactly one class of solutions, since 6|x′ x′′ − 3dy ′ y ′′ i 6|x′ y ′′ − y ′ x′′ for each two solutions
(x′ , y ′ ) and (x′′ , y ′′ ) of (20)). Hence,
√
√
√
√
1
((α + β 3d)2 (ξ + η 3d)2n + (α − β 3d)2 (ξ − η 3d)2n + 12) − 1.
36
√
According to Lemma 12, 16 (α + β 3d)2 (i.e., ( 61 (α2 + 3dβ 2 ), 31 αβ) is a fundamental solution
of x2 − 3dy 2 = 1, and so, we get (18).
If (10) is solvable in Z, we obtain (18) in similar manner. Indeed, if (x′ , y ′ ) is a solution
of (10), then c = dy ′2 − 1. All solutions of (10) are
dn =
√
√
√
√
1
yn = √ ((x0 + y0 3d)(ξ + η 3d)n + (−x0 + y0 3d)(ξ − η 3d)n ), n ≥ 0,
2 3d
where (x0 , y0 ) and (ξ, η) are fundamental solutions of (10) and (19), respectively. Further,
we have
√ 2n 1
√ 2n
√ 2
√ 2
1 1
dn =
(x0 + y0 3d) (ξ + η 3d) + (x0 − y0 3d) (ξ − η 3d) − 4 .
6 2
2
√
√
From Lemma 12, we get that 12 (x0 ± y0 3d)2 = ξ ± η 3d and again obtain (18).
Finally, we must determine in which cases dn ∈ {0, 1, 3}, since a Diophantine triple
consists of three nonzero distinct elements. Obviously, dn > 3 for n ≥ 1, so we have to see if
d0 ∈ {0, 1, 3}. Because d0 = (ξ − 2)/3, it follows that ξ ∈ {2, 5, 11} where ξ is a fundamental
solution of (19). If
8
• ξ = 2, then 5 = 3dη 2 which is not possible,
• ξ = 5, then 8 = dη 2 which implies that η = 1, d = 8 or η = 2, d = 2,
• ξ = 11, then 40 = dη 2 which implies that η = 1, d = 40 or η = 2, d = 10.
For d = 8, 40 equations (9) and (10) have no solution in Z, but (9) is solvable
√ 10
√ for d =
and (10) is solvable for d = 2. That is the reason why the element d0 in Z[ 2] and Z[ 10]
is omitted.
In Table 1 we give a list of 1 < d < 100 such that the equation√(9) is solvable. In the last
column of Table 1, the smallest c of a proper triple {1, 3, c} in Z[ d] is given.
d fund. solution of (9) fund. solution of (19) cmin
10
(2,1)
(11,2)
1763
46
(4,1)
(47,4)
15
58
(22,5)
(1451,110)
483
73
(5,1)
(74,5)
24
94
(28,5)
(2351,140)
783
Table 1: A list of 1 < d < 100 such that (9) is solvable
If the equation (10) is solvable, we obtain Table 2.
d fund. solution of (9)
2
(2,1)
6
(4,1)
17
(7,1)
18
(22,3)
22
(8,1)
34
(10,1)
38
(32,3)
41
(11,1)
54
(140,11)
57
(13,1)
66
(14,1)
82
(298,19)
86
(16,1)
89
(49,3)
97
(17,1)
fund. solution of (19) cmin
(5,2)
161
(17,4)
5
(50,7)
16
(485,66)
161
(65,8)
21
(101,10)
33
(1025,96)
341
(122,11)
40
(19601,1540)
6533
(170,13)
56
(197,14)
65
(88805,5662)
29601
(257,16)
85
(2402,147)
800
(290,17)
96
Table 2: A list of 1 < d < 100 such that (10) is solvable
9
5
Acknowledgements
The author would like to thank A. Dujella for his helpful comments.
References
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√
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√
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2000 Mathematics Subject Classification: 11D09, 11R11.
Keywords: Diophantine sets, quadratic fields, Pellian equations.
(Concerned with sequences A011922, A030063, A045899, and A051048.)
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Received August 19 2010; revised version received August 31 2010; November 17 2010.
Published in Journal of Integer Sequences, December 7 2010.
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